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3 years ago

Which situation best shows causation?

Mathematics

1 answer:

dangina [55]3 years ago

3 0

A cause and effect relationship exist between two variables, when one

variable event is made to happen by the other variable.

The situation which best shows causation is option B) The amount of time a

cell phone is used affects the charge of the battery.

Reasons:

Causation is the condition in which the occurrence of an event is caused by

or contributed to by another event. Such that the incident or observation

identified as the cause is responsible for the event, while the the

occurrence of the event depends on the cause being put in place.

Option A) A rectangle of a given width can have different dimensions of length

Option C) Milkshakes and ice cream bars can have different customer base, such that the number of milkshake and ice cream bars sold can be equal

Option D) The balls used in a soccer match are provided by the officials of the game

Therefore;

The situation which shows causation is B) The amount of time a cell phone is used affects the charge of the battery.

The option is correct because, a cell phone is powered by battery, and

therefore, the more the cell phone is used, the battery power is used up

and therefore, the charge on the battery is reduced.

Learn more here:

brainly.com/question/18404964

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Derivative of these questions

shtirl [24]

Answer: $\bold{y'=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)}$

Step-by-step explanation:

$y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\\\\\\\text{Rationalize the denominator:}\\y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\bigg(\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\bigg)\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{x+a-x-b}\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}$

$\text{Derivative of }\sqrt{x+a}\quad \rightarrow \quad (x+a)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + a)^{-\frac{1}{2}}\\\\\text{Derivative of }\sqrt{x+b}\quad \rightarrow \quad (x+b)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + b)^{-\frac{1}{2}}\\\\\\\text{Derivative of}\ \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}:\\\\= \dfrac{\dfrac{1}{2}(x + a)^{-\frac{1}{2}}+\dfrac{1}{2}(x + b)^{-\frac{1}{2}}}{(a-b)^2}$

$\text{Factor out }\dfrac{1}{2}\ \text{from the numerator:}\\\\\dfrac{(x+a)^{-\frac{1}{2}}+(x+b)^{-\frac{1}{2}}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)$

3 0

4 years ago

WILL MARK YOU BRAINLIEST PLSS HELP

matrenka [14]

H(-2) = 2(-2)^2 - 3(-2)
= 2(4) - (-6)
= 8 + 6
= 14

4 0

4 years ago

Please help ill give brainliest to the right answer

nikklg [1K]

Answer:

its the 3rd option

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

!!!!!!!!PLEASE HELP NOW !!!!!!!!!!!!!!!!!!

vampirchik [111]

Answer:

Step-by-step explanation:

You can convert the fourth square roots to $7^{\frac{1}{4}}} * 7^{\frac{1}{4}}} * 7^{\frac{1}{4}}} * 7^{\frac{1}{4}}}$ . Using the product of powers rule, we can add the four terms' exponents, resulting in $7^1$ , which is 7.